# How can we calculate the volume of a BCC Wigner-Seitz Cell? (Based on a imaginary cube)

Hello. As you can see in the picture, there’s this shape and shape’s surface consists of 6 square and 8 hexagon parts and I would like to know its volume but I don’t know where to start. The only information given is that the imaginary cube which surrounds the shape has a length of “$$a$$” for its edges, thus the imaginary cube’s volume is $$a^3$$. So, how can we find the volume of that shape? (For the edge of square in the shape, I think it has a length of $$\sqrt{a}/2$$ but I’m not sure...)

Therefore, this thing is obtained by taking an octahedron $$P$$, cut off one small pyramid at each of its six vertices. Assume the side length of $$P$$ is $$x$$, in order to make regular hexagon, the cutting points on each side is $$\frac x3$$ from vertices.
For example, let $$o$$ be a vertex of $$P$$, and $$oa, ob, oc, od$$ are the four sides from $$o$$, then take $$a'$$ on $$oa$$ with $$oa'=\frac 13 oa$$, take $$b'$$ on $$ob$$ with $$ob'=\frac 13 ob$$, take $$c'$$ on $$oc$$ with $$oc'=\frac 13 oc$$, take $$d'$$ on $$od$$ with $$od'=\frac 13 od$$, the cut away the small pyramid $$o$$-$$a'b'c'd'$$, and the new exposed face $$a'b'c'd'$$ is a square face in your object.
Now an easy computation shows the distance between two opposite vertices in $$P$$ is $$\sqrt 2 x$$. Now cut away the small pyramids, we see the distance between two opposite square faces is $$\frac 23\sqrt x$$: in fact the octahedron $$P$$ is obtained by gluing two big pyramids with side length $$x$$ together, each of them is similar to the small pyramids you cut away, the similar ratio is $$3:1$$. Anyway, this distance $$\frac 23\sqrt x$$ is just the side length (i.e. distance between two opposite faces) of your imaginary cube, therefore $$a=\frac 23 \sqrt 2 x.$$ Now the volume of the octahedron $$P$$ is $$2\cdot \frac 13 x^2\cdot\frac {\sqrt 2 x}{2}=\frac{\sqrt 2 x^3}{3}$$ (these computations are quite easy, just figure out the altitude, etc.). The volume of each of the small pyramids you cut away is $$\frac 13\cdot(\frac x3)^2\cdot (\frac x3\cdot\frac{\sqrt 2}{2})=\frac{\sqrt 2 x^3}{162}$$ ; there are six of them, therefore your object has volume $$\frac{\sqrt 2 x^3}{3}-6\cdot \frac{\sqrt 2 x^3}{162}=\frac{8\sqrt 2 x^3}{27}$$. Plug in $$x=\frac{3a}{2\sqrt 2}$$, the answer is $$\frac{a^3}{2}.$$ By the way the side length of the squares in your object is $$\frac x3=\frac{a}{2\sqrt 2}$$.